Calculate the $\text{pH}$ of a $0.01\ \text{M}$ solution of $\text{NaOH}$, assuming complete ionization.

Calculate the pH of a 0.01M solution of NaOH, assuming complete ionization.

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\text{pH}

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Concept understanding — Hydrogen Ion Concentration pH

Hydrogen Ion Concentration and pH

Imagine you have a glass of pure water. It looks simple, but inside, a tiny fraction of water molecules are constantly splitting apart and re-forming. This splitting creates two kinds of charged particles: a hydrogen ion ( $H^+$ ) and a hydroxide ion ( $OH^-$ ). In pure water, these two are perfectly balanced — there are exactly as many $H^+$ as $OH^-$ .

Now, if you add something like lemon juice (an acid), you increase the number of hydrogen ions. The balance tips: more $H^+$ than $OH^-$ . If you add baking soda (a base), you decrease $H^+$ or increase $OH^-$ , and the balance tips the other way.

The question is: how do we measure this imbalance in a simple, practical way? The numbers of $H^+$ ions are incredibly tiny — in pure water, only about 1 in every 10 million water molecules is split at any moment. Writing these numbers directly (like $0.0000001$ moles per litre) is clumsy. That's where pH comes in.

The Precise Definition

pH is a mathematical shortcut. It stands for "power of hydrogen" (from the French puissance d'hydrogène).

\text{pH} = -\log_{10}[H^+]

The logarithm base 10 does two things at once:

It compresses a huge range of numbers (from $10^{-14}$ to $10^{0}$ ) into a manageable scale of 0 to 14.
The negative sign flips the direction: higher $[H^+]$ gives a lower pH, and lower $[H^+]$ gives a higher pH.

What the Numbers Mean

$[H^+]$ (mol/L)	pH	Example
$10^{-1}$	1	Stomach acid
$10^{-3}$	3	Lemon juice
$10^{-7}$	7	Pure water (neutral)
$10^{-9}$	9	Baking soda solution
$10^{-13}$	13	Household bleach

Notice the pattern: each step of 1 in pH means a tenfold change in $[H^+]$ . A solution of pH 3 has 10 times more $H^+$ than pH 4, and 100 times more than pH 5.

The Key Insight

pH is not a measure of "how acidic" something is in a vague sense — it is a precise, logarithmic measure of the actual number of hydrogen ions present. The scale runs from 0 (most acidic, highest $[H^+]$ ) to 14 (most basic, lowest $[H^+]$ ), with 7 being neutral. …

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$\text{NaOH}$ is a strong base and ionizes completely, so $[\text{OH}^-] = 0.01\ \text{M} = 10^{-2}\ \text{M}$ directly. $\text{pOH} = -\log_{10}(10^{-2}) = 2$ . Using $\text{pH} + \text{pOH …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 25 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. NaOH(aq), HCl(aq) and NaCl(aq) have concentration of 10^-3 M each. Their pH will be respectively$
(a) 10, 6, 2
(b) 11, 3, 7
(c) 10, 3, 7
(d) 10, 4, 7
›Reveal solutionSolution

NaOH and HCl are both strong electrolytes that fully dissociate, giving [OH-] or [H+] equal to the stated concentration; NaCl is a neutral salt (from a strong acid and strong base) and does not affect pH.

NaOH (strong base), 10^-3 M:

Fully dissociates: [OH-] = 10^-3 M

pOH = -log(10^-3) = 3

pH = 14 - pOH = 14 - 3 = 11

HCl (strong acid), 10^-3 M:

Fully dissociates: [H+] = 10^-3 M

pH = -log(10^-3) = 3

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The pH value of gastric juice is -$
(a) -3
(b) -1.2
(c) -2
(d) -4
›Reveal solutionSolution

The pH of gastric juice is commonly quoted as approximately 1.2 (highly acidic); of the given options, (b) -1.2 matches this value in magnitude.

Gastric juice, secreted by the stomach's gastric glands, contains hydrochloric acid (HCl) at a concentration that gives it a strongly acidic pH, essential for activating pepsin and digesting food. NCERT's own table of pH of some common fluids lists gastric juice at pH ≈ 1.2.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. A solution with a pH value less than 7 is:$
(a) Basic
(b) Acidic
(c) Saline
(d) Neutral
›Reveal solutionSolution

pH < 7 = acidic, pH = 7 = neutral, pH > 7 = basic, on the standard 0-14 pH scale.

pH is defined as pH = -log10[H3O+]. A lower pH corresponds to a higher concentration of hydronium (H3O+) ions. Pure water at 25°C has pH = 7 and is neutral. Solutions with pH less than 7 have a higher H+ concentration than pure water and are called acidic; solutions with pH greater than …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The pH of an aqueous solution is 4.0, the value of its pOH would be$
(a) 4.0
(b) 6.0
(c) 8.0
(d) 10.0
›Reveal solutionSolution

pOH = 14 − pH = 14 − 4 = 10.

For any aqueous solution at 25°C, the ionic product of water gives pH + pOH = 14. …

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CBSE 2026Set ANNUAL1 mark
$Q. What is the value of Kw at 25 C ?$
›Reveal solutionSolution

The ionic product of water Kw at 25 C is 1.0 x 10^-14.

Water self-ionises: H2O ⇌ H+ + OH-. The ionic product Kw = [H+][OH-]. At 25 C pure water has [H+] = [OH-] = 1.0 x 10^-7 mol L^-1, so:

Kw = (10^-7)(10^-7) = 1.0 x 10^-14 mol^2 L^-2. …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. With increase in temperature, ionic product of water$
(a) Increases
(b) Decreases
(c) Remains unaffected
(d) Zero
›Reveal solutionSolution

Kw (the ionic product of water) increases with rising temperature.

Water's self-dissociation (2H2O ⇌ H3O+ + OH-) is an endothermic process. By Le Chatelier's principle, increasing temperature shifts an endothermic equilibrium further towards the products, increasing the concentrations of H3O+ and OH-, a …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following has lowest pH value?$
(a) 1M HCl
(b) 1M NaOH
(c) 1M H2SO4
(d) 1M C2H5OH
›Reveal solutionSolution

1M H2SO4 has the lowest pH because, being diprotic, it furnishes roughly twice the [H+] of 1M HCl.

HCl and H2SO4 are both strong acids, but HCl is monoprotic (1M HCl gives [H+] ≈ 1M, pH = 0) while H2SO4 is diprotic (1M H2SO4 gives [H+] ≈ 2M, since both protons ionise essentially completely, pH = -log(2) ≈ -0.3). NaOH is a strong base (high pH, ~14), and ethanol (C2H5OH) i …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If 50 ml of 0.1 M HCl and 50 ml of 0.2 M NaOH are mixed then the pH of the resulting solution will be$
(a) 1.30
(b) 4.2
(c) 12.70
(d) 11.70
›Reveal solutionSolution

Mixing 50 mL of 0.1M HCl with 50 mL of 0.2M NaOH leaves excess NaOH, giving pH = 12.70.

Moles of HCl = 0.050 L x 0.1 mol/L = 0.005 mol.

Moles of NaOH = 0.050 L x 0.2 mol/L = 0.010 mol.

HCl + NaOH → NaCl + H2O consumes 0.005 mol of each, leaving excess NaOH = 0.010 - 0.005 = 0.005 mol.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The value of pH of 0.1 M NaOH solution is equal to$
(a) 13
(b) 1.2
(c) 5.0
(d) 1.0
›Reveal solutionSolution

NaOH is a strong base (100% dissociated); its [OH-] equals its molarity, so pOH = 1 and pH = 14 - pOH = 13.

Step 1 — Dissociation: NaOH is a strong base, meaning it dissociates completely in water:

NaOH -> Na+ + OH-

Since dissociation is 100%, the hydroxide ion concentration equals the given molarity:

[OH-] = 0.1 M = 10^-1 M

Step 2 — Calculate pOH:

pOH = -log10[OH-] = -log10(10^-1) = 1

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If concentration of hydrogen ion [H+] changes by a factor of 100, then the value of pH changes by:$
(a) one unit
(b) 2 unit
(c) 10 unit
(d) 100 unit
›Reveal solutionSolution

Since pH is a logarithmic (base 10) scale, multiplying [H $^+$ ] by 100 shifts pH by exactly $\log_{10}(100) = 2$ units.

By definition:

$pH = -\log_{10}[H^+]$

Let the initial concentration be $[H^+]_1$ and the new concentration be $[H^+]_2 = 100 \times [H^+]_1$ (changed by a factor of 100).

$pH_1 = -\log[H^+]_1$

$pH_2 = -\log(100 \times [H^+]_1) = -\log(100) - \log[H^+]_1 = -2 + pH_1$

$\Delta pH = pH_1 - pH_2 = 2$

…

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CBSE 2025Set hz1 markMCQ
$Q. Select the correct one: The hydronium ion concentration i.e. [H3O+] of a solution is 2x10^-5 M, its pH is:$
(a) 5
(b) 7
(c) 4.7
(d) 3
›Reveal solutionSolution

pH = -log[H3O+]; substituting [H3O+] = 2x10^-5 M gives pH = 4.7.

By definition: pH = -log10[H3O+]

Given [H3O+] = 2 x 10^-5 M.

pH = -log(2 x 10^-5)

= -[log 2 + log 10^-5] …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The pH of 10^-4 M NaOH solution is$
(a) 4
(b) 10
(c) 12
(d) 2
›Reveal solutionSolution

10⁻⁴ M NaOH has pOH = 4, so pH = 10.

NaOH is a strong base, fully dissociated, so $[OH^-] = 10^{-4}$ M.

…

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Q.Calculate the pH\text{pH}pH of a 0.01 M0.01\ \text{M}0.01 M solution of NaOH\text{NaOH}NaOH, assuming complete ionization.

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Q.Calculate the $\text{pH}$ of a $0.01\ \text{M}$ solution of $\text{NaOH}$ , assuming complete ionization.

Unlock everything free for 14 days